Let the line x - y = 4 intersect the circle C | vedclass

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(A) The circle $C$ has center $O(4, -3)$ and radius $r = 3$.The line $L: x - y - 4 = 0$ intersects the circle at $Q$ and $R$.For $PQ = PR$, $P$ must l

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) The circle $C$ has center $O(4, -3)$ and radius $r = 3$.The line $L: x - y - 4 = 0$ intersects the circle at $Q$ and $R$.For $PQ = PR$, $P$ must lie on the perpendicular bisector of the chord $QR$.The perpendicular bisector of any chord passes through the center of the circle.The slope of line $L$ is $1$, so the slope of the perpendicular bisector is $-1$.The equation of the perpendicular bisector passing through $(4, -3)$ is $y - (-3) = -1(x - 4)$, which simplifies to $x + y = 1$.Point $P(\alpha, \beta)$ lies on the circle and the line $x + y = 1$, so $\beta = 1 - \alpha$.Substituting into the circle equation: $(\alpha - 4)^2 + (1 - \alpha + 3)^2 = 9$.$(\alpha - 4)^2 + (4 - \alpha)^2 = 9 \implies 2(\alpha - 4)^2 = 9 \implies (\alpha - 4)^2 = 4.5$.Also, $6\alpha + 8\beta = 6\alpha + 8(1 - \alpha) = 8 - 2\alpha$.From $(\alpha - 4)^2 = 4.5$, $\alpha - 4 = \pm \frac{3}{\sqrt{2}}$, so $\alpha = 4 \pm \frac{3}{\sqrt{2}}$.Then $8 - 2\alpha = 8 - 2(4 \pm \frac{3}{\sqrt{2}}) = 8 - 8 \mp 3\sqrt{2} = \mp 3\sqrt{2}$.Squaring this, $(6\alpha + 8\beta)^2 = (\mp 3\sqrt{2})^2 = 9 	imes 2 = 18$.

Let the line $x - y = 4$ intersect the circle $C : (x - 4)^2 + (y + 3)^2 = 9$ at the points $Q$ and $R$. If $P(\alpha, \beta)$ is a point on $C$ such that $PQ = PR$, then $(6\alpha + 8\beta)^2$ is equal to . . . . . .

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